Dc Machines Practice Questions

Field Control: Fieldcontrol provides constant power drive. Since the speed is inversely proportional to flux/pole $\left ( \omega \propto \frac{1}{\phi } \right )$ field can not be increased; It can only be weakended. So this control is suitable for speed control above base speed as torque is directly proportional to flux/pole $(T \propto \alpha )$ for a given armature current. So, this control provides torque below base torque. Armature Control: It provides constant-torque drive. By keeping th field current at maximumvalue full load torque can be obtained at full load armature current at all speeds. This control provides speed below base speed.

In series motor, Back emf, $E=v-I(R_a+R_{se})$ $\text{If},\;\;(R_a+R_{se})=R$ $\text{them,}\;\; E=V-IR$ $\text{where,}\;\; E=K_n \phi N$ $\;\;\;\;\;N=\frac{V-IR}{K_n \phi }$ In series motor, $\phi \propto I$ $\therefore$ At constant power load $E \times I=T \times w=const$ $\text{where,} \;\; T=K_a\phi I=K_a I^2$ If ' $w_n$ ' decrease by (1/4) of its rated value, then torque will increase 4 times to maintain power contant. $\therefore \;\; T\propto I^2$ $\therefore$ current increase 2 times so that supply voltage approximately brought down to 0.5pu.

In DC machine, field is provided at stator and armature is provided at rotor. Because of commutator and slip-rings, AC rotor winding current is converted inot dc. IN synchronous machine, it is difficult to provide armature in rotor because of many factor like insulation no of slip rings, rating of slip rings, high rating of armature. Therefore, armature is provided in stator and field in rotor Induction machines are singlyfed machine i.e. they draw excitation from the same supply. So no external field supply is required.

All calculation are done in p.u. system. Motor is connected across rated voltage V= 1 pu. Rated armature current flows in both motor and generator $I_g=I_m=1 pu$ Armature resistance of both motor and generator $R_m=R_g=0.02 pu$ Back iemf in motor $=E_b=V-I_mR_m$ $=1-1 \times 0.02 =0.98 pu$ Mechanical power output of the motor $=E_bI_m=0.98 \times 1=0.98 pu$ This power is given to the generator and armation reaction and mechanical losses are negligible. Input power of generator = Output power of motor $E_bI_g=E_bI_m E_g=E_b=0.98pu$ Terminal voltage of the generator $V_g=E_g-I_gR_g =0.98-1 \times 0.02=0.96pu$ $R_L=\frac{V_g}{I_g}=\frac{0.96}{1}=0.96\Omega$

At saturation , linear characteristic are observed.

$P=\frac{2 \pi N}{60}T=\text{constant}$ $xy=C$ (Rectangular Hyperbola)